Arescue pilot drops a survival kit while her plane is flying at an altitude of 1.5 km with a forward velocity of 100.0 m/s. if air friction is disregarded, how far in advance of the starving explorer’s drop zone should she release the package?

Given: Initial Velocity Vi = 100 m/s; Altitude y = 1.5 Km; Time t = ?

The aircraft is moving horizontally on x direction

Formula to follow: x = VicosΘt   but Time t is unknown. 

Solving for Time "t" formula will be y = VisineΘt + 1/2 gt² 

Since sineΘt = 0 therefore  y = 1/2gt² Solve for t

t = √2y/g  t = √2(1500 m)/9.8m/s²)   t = 31.24 s

Now solving for distance

x = VicosΘt  x = (100 m/s) (31.24 s) x =  3,124 m or 3.124 Km

 

What even is this. where is the question